TMJC 9758 2025 Prelim P2 Solutions

TMJC/2025 JC2 Preliminary Examination Markers Comments/H2 Math (9758/02)/Math Dept Page 1 of 17 2025 H2 MATH (9758/02) JC 2 PRELIMINARY EXAMINATION Qn Solution 1 Equations and Inequalities (a) ( ) ( ) ( ) ( )( ) 2 2 2 2 2 2 2 2 2 2 2 5 12 5 3 5 102 5 3 5 2 5 3 02 5 3 5 2 5 3 02 5 3 2 4 2 02 5 3 2 2 1 02 5 3 21 10, , 33 2 1 2 x xx x xx x x x xx x x x xx xx xx xx xx x xxxx + − + + + −− + + + − − + + − + + + + − − − + + −+ − + + −+ − + + −  − −+ 1 or 32xx−  (b) 2 cos 5 12cos 5cos 3 x xx + − + + Replace x by cos x : 1cos or cos 3 (Rej. 1 cos 1)2x x x−  −   2 3 x    0 x y 1 + + 3 − −

TMJC/2025 JC2 Preliminary Examination Markers Comments/H2 Math (9758/02)/Math Dept Page 2 of 17 Qn Solution 2 Maclaurin Series (a) Since the curve ( ) 2f a b c  = + + passes through ( )0,1 , 1c= Since the curve ( ) 2f a b c  = + + passes through 31,32 − , 2 3 3 1 1 3 1 13 3 2 3 3 2a b a b   − + − + =  − =−          …….(1) ( )f ' 2 ab=+ Since the turning point is at 31,32 − , ( )f ' 0 = ( ) 320 3 23 03 23 23 . ab ab ba − + =  += =   − Sub (2) into (1): 1 3 1 3 3 2 1 3 2 3 1 3 3 3 2 1 1 3 3 2 2 ab aa aa − =− − =− − =−  = 2 3 2 3 3 33 3 2ba = = =  3 , 3, 12a b c= = = (b) ( )( ) 2 2 2 2 cos 6BC AB AC AB AC = + − +  ( ) ( )( ) 221 3 2 1 3 cos cos sin sin 66 314 2 3 cos sin22 4 3cos 3 sin    = + − −  = − −  = − +

TMJC/2025 JC2 Preliminary Examination Markers Comments/H2 Math (9758/02)/Math Dept Page 3 of 17 ( ) 2 2 2 when is small ) 4 3 1 3 2! 34 3 3 2 313 2 f (Shown        − − + = − + + = + + = (c) ( ) 2 1 22 2 22 22 2 313 2 313 2 11 11 3 3 221 3 32 2 2! 2 3 3 113 2 4 8 331 (Shown)28 BC            + +  + +  −     + + + +        + + − = + +

TMJC/2025 JC2 Preliminary Examination Markers Comments/H2 Math (9758/02)/Math Dept Page 4 of 17 Qn Solution 3 Graphing Techniques, Transformation of Curves (a) (b) x y O x y O

TMJC/2025 JC2 Preliminary Examination Markers Comments/H2 Math (9758/02)/Math Dept Page 5 of 17 Qn Solution 4 Differential Equations (a) 22 2 dd 0.05dd PP tt =−  Substituting d d Pv t= , we get 2d 0.05d v vt =− Using separable variables, 2 1 d 0.05 dvtv =− 1 0.05 'tcv− =− + where 'c is an arbitrary constant 1 0.05 ' 20 v tc tC = − = + where 20 'Cc=− is a constant (shown) (b) 20v tC= + Given 0 and 10Pv== when 0,t= 2010 2 CC=  = 20 2v t= + d 20 d2 P tt= + 20ln 2P t D = + + where D is a constant when 0,t= 0,P= 0 20ln 2 20ln 2 D D  = + =− 20ln 2 20ln 2 220ln since 02 Pt t t = + − +=  (c) When 30 230, 20ln 55.5 2tP += = =  (3 s.f)

TMJC/2025 JC2 Preliminary Examination Marker